Question

How do you solve $2h-8=h+17?$

Answer 1

Hint: Firstly, we learn what linear equation is in 1 variable term. We use a hit and trial method to find the value of ‘h’. In this method we put the value of ‘h’ one by one by hitting arbitrary values and looking for needed values. Another method is to apply algebra. We subtract terms to get to our final term and get our required solution.

Complete answer:
We are given that we have $2h-8=h+17$. We are asked to find the value of ‘h’, or we are asked how we will be able to solve this expression.
We learn about the equation on one variable. one variable simply represents the equation that has one variable (say x, y, or z) and another one constant.
For Example:
$x+2=4,2-x=2,2x,2y$ Etc.
Our equation $2h-8=4+17$ also has just one variable ‘h’.
We have to find the value of ‘h’ which will satisfy our given equation.
Firstly we try by the method of hit and trial. In which we will put a different value of ‘h’ and take which one fits the solution correctly.
Let $h=0$
We put $h=0$ in $2h-8=h+17$, we get –
$2\times 0-8=0+17$
$-8=17$
Which is not true, so,
$h=0$ is not the solution of this.
Let $h=1$
We put $h=1$ in $2h-8=h+17$, we get –
$\begin{align}
  & 2\times 1-8=1+17 \\
 & 2-8=18 \\
 & -6=18 \\
\end{align}$
Which is not true, so
$h=1$ is also not the solution of this.
Let $h=20$
We put $h=20$ in $2h-8=h+17$, we get –
$\begin{align}
  & 2\times 20-8=20+17 \\
 & 40-8=37 \\
 & 32=37 \\
\end{align}$
Which is not true, so
$h=20$ is also not the solution of this.
If we put $h=25$ in $2h-8=h+17$, we get –
$\begin{align}
  & 2\times 25-8=25+17 \\
 & 50-8=42 \\
 & 42=42 \\
\end{align}$
Which is true, so
$h=25$ is the solution.
Another way to solve this is to use the algebraic operation, we use multiplication, division, addition between variables and constants to simplify and solve.
We have –
$2h-8=h+17$
Subtract ‘h’ on both sides, we get –
$\begin{align}
  & 2h-h-8=h-h+17 \\
 & \Rightarrow 4-8=17 \\
\end{align}$
Adding ‘8’ on both sides, we get –
$\begin{align}
  & h-8+8=17+8 \\
 & h=25 \\
\end{align}$
So,
$h=25$ is the solution of this problem.

Note: Remember that, we cannot add the variable to the constant. Usual mistakes like this where one adds constants with variables usually happen.
For example: $3x+6=9x$ , here one added ‘6’ with 3 of x made it 9x, this is wrong, we cannot add constant and variable at once. Only the same variables are added to each other.